Answer:
P(A∣D) = 0.667
Step-by-step explanation:
We are given;
P(A) = 3P(B)
P(D|A) = 0.03
P(D|B) = 0.045
Now, we want to find P(A∣D) which is the posterior probability that a computer comes from factory A when given that it is defective.
Using Bayes' Rule and Law of Total Probability, we will get;
P(A∣D) = [P(A) * P(D|A)]/[(P(A) * P(D|A)) + (P(B) * P(D|B))]
Plugging in the relevant values, we have;
P(A∣D) = [3P(B) * 0.03]/[(3P(B) * 0.03) + (P(B) * 0.045)]
P(A∣D) = [P(B)/P(B)] [0.09]/[0.09 + 0.045]
P(B) will cancel out to give;
P(A∣D) = 0.09/0.135
P(A∣D) = 0.667
Lets say the 3x3 Matrix is
M = [1 5 2 ]
[1 1 7 ]
[0 -3 7 ]
We apply the Gauss-Jordan elimination method
(Procedure and result shown in the image below)
-4 + 20- 100 + 500 - 2500 + 12,500 - 62,500 + 312,500 = 260,416